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Homework Help: Microwave reflection problem

  1. Sep 2, 2013 #1
    1. The problem statement, all variables and given/known data
    A plastic plate with refraction index of 1.5 is placed in the interior of a micro-wave oven which operates with a frequency of 2.5 x 10^9 Hz. If the micro-waves are perpendicular to the surface of the plate, what is the minimum thickness of the plate so that the maximum reflection of micro-waves occurs?

    2. Relevant equations

    3. The attempt at a solution
    I honestly don't even understand the premises of the problem.. I feel like the maximum reflection of micro-waves will occur when some destructive interference is minimized. Right? But what are the interfering waves in this case? Are some waves reflecting out of the top surface of the plate and some go all the way through and reflect at the bottom surface? In that case, why is that happening?
  2. jcsd
  3. Sep 3, 2013 #2


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    Homework Helper

    Yes. The waves reflected from the top of the plate interfere with those, reflected from the bottom surface of the plate. Microwaves are electromagnetic waves, like light waves. Incident upon a boundary between two different media, part of the beam is reflected, some refracted and travel in the other medium, and reaching the second interface, partly reflected again. It is the same as with light waves. What is the condition for maximum reflection?

  4. Sep 3, 2013 #3
    Okay so..

    The waves that go all the way through and reflect at the bottom of the plate have to travel a distance ## 2d ## inside the plate, where ## d ## is the thickness of the plate. The minimum thickness so that those waves won't cause destructive interference is ## \frac{1}{2}\lambda_{plate} ##. Because if ## 2d=\frac{1}{2}\lambda_{plate} ## the waves that bounce back will have inverted phase.

    And if ## \frac{\lambda_{plate}}{\lambda_{air}}=\frac{n_{air}}{n_{plate}} ## from snell's law, then ## \frac{\lambda_{plate}}{\lambda_{air}}=\frac{1}{1.5} ##. And ## v=f\lambda ##, so ## \lambda_{air}=\frac{c}{f}=\frac{3x10^8}{2.5x10^9} ##.

    Therefore ## \lambda_{plate}=\frac{3x10^8}{1.5*2.5^10^9}=0.08m=8cm##

    So the minimum thickness ## d=\frac{1}{4}\lambda_{plate} ## is ## \frac{1}{4}8cm=2cm ##

    But I still don't get why some waves bounce back at the top and some bounce at the bottom.. Is it probabilistic??
  5. Sep 3, 2013 #4


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    You have to imagine the incident wave like water waves coming in to the shore. They never stop.
    Part of the wave reflects, the other part enters into the plate, reaches the bottom, and part of it reflects again. The intensity of the wave decreases at every reflection. That back-reflected wave reaches the first surface and part of it goes into the air again. That wave would interfere with the wave, just arrived and reflected back directly from the first surface. But this, directly reflected wave changes phase at reflection: Its phase changes by pi. If the incident wave had maximum, the reflected one has minimum.
    When the wave reflected from the back surface has the same phase as the first one, there is constructive interference, the reflection is high.
    In order to get phase opposite to that of the incident wave, the refracted beam has to travel λ/2 when traversing the plate forward and back. Your solution is correct.

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